Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 26, 2007, 27–31

© K. Fukuyama and R. Kondo

Abstract. The Riesz-Raikov sums $\sum f\left({\theta }^{k}x\right)$ are recurrent in most cases.

1. Introduction

In the theory of lacunary series, some probabilistic limit theorems are proved for gap series $\sum f\left(n_{k}x\right)$. Among these, we focus on the recurrence property.

Hawkes [5] proved that ${\left\{{\sum }_{k=1}^{N}exp\left(i{n}_{k}x\right)\right\}}_{N\in N}$ is dense in complex plain for a.e. $x$ assuming very strong gap condition $\sum n_k∕n_{k+1}<\infty$. Anderson and Pitt [1] used the theory of Bloch function and weaken the gap condition to $n_{k+1}∕n_k\to \infty$ or $n_k=a^k$, where $a\ge 2$ is an integer. These results imply that ${\left\{{\sum }_{k=1}^{N}cos{n}_{k}x\right\}}_{N\in N}$ is dense in the real line. As to this one-dimensional recurrence, Ullich, Grubb and Moore [10], [4] succeeded in weakening the gap condition to the Hadamard’s $n_{k+1}∕n_k>q>1$. The purpose of this paper is to show that their real analytic proof is also effective for general gap series $\sum f\left(n_{k}x\right)$ where $f$ is not necessarily analytic function, which seems difficult to treat by the method of Anderson and Pitt.

We assume that $f$ is a real-valued function on $R$ with period $1$ satisfying

for some $\alpha >0$ and $M>0$.

We first consider the Riesz-Raikov sum $\sum f\left({\theta }^{k}x\right)$, where $\theta >1$ is a real number. In the case when the condition

is satisfied, partial sum of $\sum f\left({\theta }^{k}x\right)$ always reduced to sum of at most $2\rho$ terms, and may not be recurrent on the whole line. Except for this trivial case, we have the recurrence property.

By noting the case (2) we see that the Hadamard’s gap condition is not enough to assure the recurrence of $\sum f\left(n_{k}x\right)$. It is, by the way, possible to prove the recurrence by assuming a stronger gap condition:

We prove these results by the method of Ullich, Grubb and Moore, with the help of the mixing central limit theorem. We here introduce that notion. Let $\lambda$ be the Lebesgue measure on $I=\left[0,1\right]$ and $\left\{g_n\right\}$ be a sequence of measurable functions on $I$. We say that $\left\{g_n\right\}$ obeys the mixing central limit theorem with limiting variance $v$ if the probability measure $\frac{1}{\lambda \left(E\right)}\lambda \left(\left\{x\in E\mid g_n\left(x\right)\in \cdot \right\}\right)$ on $R$ converges weakly to the normal distribution $N_{0,v}$ with mean $0$ and variance $v$ for any $E\subset I$ with positive measure.

Study of the Riesz-Raikov sum has long history [7], [6], [8], and we [2] have proved the mixing central limit theorem holds for ${\sum }_{k=1}^{n}f\left({\theta }^{k}x\right)∕\sqrt{n}$, where $\theta >1$. The limiting variance $v$ is given by

if ${\theta }^r\notin Q$ for all $r\in N$, and

if $\theta =\sqrt[r]{p∕q}$ where $r=min\left\{k\in N\mid {\theta }^k\in Q\right\}$ and $p$, $q\in N$. Here $v$ is always non-negative, and is equal to $0$ if and only if (2) holds.

As to the case of Theorem 2, we have the mixing central limit theorem with $v={\int \; <!--nolimits-->}_0^1{f}^2\left(x\right)dx$ which was proved by Takahashi [9] by assuming that ${\beta }_k$ are integers. We can easily drop the last condition in the same way as [2].

Lastly, we present another case when the mixing central limit theorem is proved. Let $\theta >1$, real-valued functions $f_1$, …, $f_L$ on $R$ satisfy (1), and $p_1$, …, $p_L$ be polynomials satisfying $p_m\left(\infty \right)=\infty$ and $\left(p_{m+1}-p_m\right)\left(\infty \right)=\infty$. Then ${\sum }_{k=1}^n{\prod }_{m=1}^L{f}_m\left({\theta }^{p_m\left(k\right)}x\right)∕\sqrt{n}$ obeys the mixing central limit theorem. Limiting variance is given as follows: If at lease one of the $p_m\left(k\right)$ is not linear, then

When all $p_m$ are linear, i.e., $p_m\left(x\right)=a_{m}x+b_m$, if there exists $m$ such that ${\theta }^{a_{m}r}\notin Q$ for all $r\in N$, then $v$ is given as above. If ${\theta }^{a_{m}r}=p_m∕q_m$ ($m=1$, …, $L$), then

By the help of this result, we can prove

2. Proof of the Theorems

To verify our results, it is sufficient to prove the proposition below:

We use the lemma below proved by Ullich [10], [4].

We follow the proof given by Grubb and Moore [4]. Take $M$ large enough to satisfy both (1) and $\mid f_1\left(x\right)\mid$, …, $\mid f_L\left(x\right)\mid \le M$ for all $x$. Put $g_k\left(x\right)={\prod }_{m=1}^L{f}_m\left({\beta }_{m,k}x\right)$. We have $\mid g_k\left(x+h\right)-g_k\left(x\right)\mid \le M^L\mid h{\mid }^{\alpha }{\sum }_{l=1}^L{\beta }_{l,k}^{\alpha }$ by $\mid {\xi }_1\dots {\xi }_m-{\eta }_1\dots {\eta }_m\mid \le {\sum }_{l=1}^L\mid {\xi }_1\dots {\xi }_{l-1}\left({\xi }_l-{\eta }_l\right){\eta }_{l+1}\dots {\eta }_m\mid$. By applying ${\sum }_{k=1}^n{\beta }_{l,k}^{\alpha }\le {\beta }_{l,n}^{\alpha }\left(1+1∕q^{\alpha }+1∕q^{2\alpha }+\cdots \right)={\beta }_{l,n}^{\alpha }∕\left(1-1∕q^{\alpha }\right)$, we have $\mid S_n\left(x+h\right)-S_n\left(x\right)\mid \le C\mid h{\mid }^{\alpha }{max}_{l=1}^L{\beta }_{l,n}^{\alpha }$, where $C=L{M}^L∕\left(1-1∕q^{\alpha }\right)$.

Let us take small $\varepsilon >0$ satisfying $c={\left(\varepsilon ∕C\right)}^{1∕\alpha }<1∕2$. Put $E_n=\left\{x\in I\mid S_n\left(x\right)\ge a,S_{n+1}\left(x\right)\le a\right\}$, $F_n=\left\{x\in I\mid S_n\left(x\right)or{S}_{n+1}\left(x\right)\in \left(a-\varepsilon ,a+\varepsilon \right)\right\}$, and $G_{\pm }=\left\{x\in I\mid \pm S_n\left(x\right)\ge \pm af.e.\right\}$. If $\lambda \left(G_{+}\right)>0$ we have

where the right hand side tends to $N_{0,v}\left(-\mid a\mid ,\infty \right)<1$ by the mixing central limit theorem, which contradicts with a definition of $G_{+}$. In the same way we have $\lambda \left(G_{-}\right)=0$, and therefore we have proved $x\in E_n$ i.o. for a.e. $x$.

Let $x\in E_n$. Defining $l_0$ and ${\delta }_n$ by ${max}_{l=1}^L{\beta }_{l,n+1}={\beta }_{l_0,n+1}=1∕{\delta }_n$, we have ${max}_{l=1}^L{\beta }_{l,n}^{\alpha }\le {max}_{l=1}^L{\beta }_{l,n+1}^{\alpha }=1∕{\delta }_n^{\alpha }$. Let $J=\left(x-{\delta }_n∕2,x+{\delta }_n∕2\right)$.

Firstly, we assume that there exists an $x_0\in J$ such that $S_n\left(x_0\right)=a$. If $\mid y-x_0\mid <c{\delta }_n$, we have $\mid S_n\left(y\right)-a\mid \le C\mid y-x_0{\mid }^{\alpha }∕{\delta }_n^{\alpha }<\varepsilon$. Noting $c<1∕2$, we see that $\left(x_0-c{\delta }_n,x_0\right)$ or $\left(x_0,x_0+c{\delta }_n\right)$ is contained in $J\cap F_n$, and thereby $\lambda \left(J\cap F_n\right)\ge c{\delta }_n=c\lambda \left(J\right)$.

Secondly we assume that $S_n>a$ on $J$. Since $J$ contains a period of $f_1\left({\beta }_{l_0,n+1}\cdot \right)$, $J$ contains its zero $x_1$. By ${\prod }_{m=1}^L{f}_m\left({\beta }_{m,n+1}{x}_1\right)=0$, we have $S_{n+1}\left(x_1\right)=S_n\left(x_1\right)>a$. Because of $S_{n+1}\left(x\right)\le a$, we have $x_2\in J$ between $x$ and $x_1$ such that $S_{n+1}\left(x_2\right)=a$. Therefore, if $\mid y-x_2\mid \le c{\delta }_n$, we have $\mid S_{n+1}\left(y\right)-a\mid \le C\mid y-x_2{\mid }^{\alpha }∕{\delta }_n^{\alpha }<\varepsilon$, and thereby $\left(x_2-c{\delta }_n,x_2\right)$ or $\left(x_2,x_2+c{\delta }_n\right)$ is contained in $J\cap F_n$, and hence $\lambda \left(J\cap F_n\right)\ge c{\delta }_n=c\lambda \left(J\right)$.

We have verified the assumption of Lemma and proved $x\in F_n$ i.o. a.e. $x$.

References

Department of Mathematics, Kobe University, Rokko, Kobe, 657-8501, Japan

E-mail address: fukuyama@math.kobe-u.ac.jp

Received January 12, 2007